Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 119282 by mnjuly1970 last updated on 23/Oct/20

... nice calculus... evaluate:: I:= ∫_0 ^( 1) li_2 (1−x^2 )dx=?? .m.n.1970.

...nicecalculus...evaluate::I:=∫01li2(1−x2)dx=??.m.n.1970.

Answered by mindispower last updated on 23/Oct/20

by part,(d/dx)li_2 (x)=−((ln(1−x))/x)⇒ =[xli_2 (1−x^2 )]_0 ^1 −∫_0 ^1 ((2x^2 ln(x^2 ))/(1−x^2 ))dx =−∫_0 ^1 ((2(x^2 −1+1)ln(x^2 ))/(1−x^2 )) =2∫_0 ^1 ln(x^2 )−2∫_0 ^1 ((ln(x^2 ))/(1−x^2 ))dx 4[xln(x)−x]_0 ^1 −4∫_0 ^1 Σx^(2k) ln(x) −4+4Σ_(k≥0) (1/((2k+1)^2 ))=−4+4.(3/4)ζ(2)=−4+(π^2 /2)

bypart,ddxli2(x)=−ln(1−x)x⇒=[xli2(1−x2)]01−∫012x2ln(x2)1−x2dx=−∫012(x2−1+1)ln(x2)1−x2=2∫01ln(x2)−2∫01ln(x2)1−x2dx4[xln(x)−x]01−4∫01Σx2kln(x)−4+4∑k⩾01(2k+1)2=−4+4.34ζ(2)=−4+π22

Commented by mnjuly1970 last updated on 23/Oct/20

perfect mr power thank you...

perfectmrpowerthankyou...

Commented by mindispower last updated on 23/Oct/20

withe pleasur

withepleasur

Terms of Service

Privacy Policy

Contact: info@tinkutara.com